2020ICPC上海 E.The Journey of Geor Autumn

题目链接

题目描述
Once upon a time, there was a witch named Geor Autumn, who set off on a journey across the world. Along the way, she would meet all kinds of people, from a country full of ICPC competitors to a horse in love with dota—but with each meeting, Geor would become a small part of their story, and her own world would get a little bit bigger.

Geor just arrived at the state of Shu where people love poems. A poem is a permutation (a1,…,an){(a_1,\ldots, a_n)}(a1,…,an) of [n]{[n]}[n]. ((a1,…,an)((a_1,\ldots, a_n)((a1,…,an) is a permutation of [n]{[n]}[n] means that each ai{a_i}ai is an integer in [1,n]{[1,n]}[1,n] and that a1,…,an{a_1,\ldots, a_n}a1,…,an are distinct.) One poem is good if for all integer i{i}i satisfying i>k{i> k}i>k and i≤n{i\le n}i≤n, ai>min⁡(ai−k,…,ai−1){a_i>\min(a_{i-k}, \ldots, a_{i-1})}ai>min(ai−k,…,ai−1). Here min⁡(ai−k,…,ai−1){\min(a_{i-k}, \ldots, a_{i-1})}min(ai−k,…,ai−1) denotes the minimum value among ai−k,…,ai−1{a_{i-k}, \ldots, a_{i-1}}ai−k,…,ai−1 .

Help Geor calculate how many good poems there are, given n{n}n and k{k}k. To avoid huge numbers, output the answer modulo 998244353{998244353}998244353.

输入描述:
The first line contains two integers nnn and kkk separated by a single space (1≤n≤1071\le n\le 10^71≤n≤107
, 1≤k≤1071\le k\le 10^71≤k≤107 ).

输出描述:
Output only one integer in one line—the number of good poems modulo 998244353{998244353}998244353.

示例1
输入

1 1

输出

设 dpidp_idpi 为长度为 iii 的序列满足条件的数量。
显然要想满足条件当前序列最小的元素应该放在前 kkk 个位置，假设放在 jjj ，则位置 jjj 之前可以放任何数，即有 Ai−1j−1A_{i-1}^{j-1}Ai−1j−1 种方法，位置 jjj 之后的序列可以看做是一个新序列，需要满足条件，因此有 dpi−jdp_{i-j}dpi−j 种。因此 dpi=∑j=1min⁡(i,k)Ai−1j−1dpi−jdp_i=\sum\limits_{j=1}^{\min(i,k)}A_{i-1}^{j-1}dp_{i-j}dpi=j=1∑min(i,k)Ai−1j−1dpi−j，即 dpii!=1i∑j=1min⁡(k,i)dpi−j(i−j)!\frac{dp_i}{i!}=\frac{1}{i}\sum\limits_{j=1}^{\min(k,i)}\frac{dp_{i-j}}{(i-j)!}i!dpi=i1j=1∑min(k,i)(i−j)!dpi−j 。
令 pi=dpiip_{i}=\frac{dp_i}{i}pi=idpi ，则 pi=1i∑j=1min⁡(k,i)pjp_i=\frac{1}{i}\sum\limits_{j=1}^{\min(k,i)}p_jpi=i1j=1∑min(k,i)pj 。最终答案为 n!pnn!p_nn!pn 。时间复杂度为 O(n)O(n)O(n) 。

#include<bits/stdc++.h>using namespace std;
const int N = 1e7 + 10, MOD = 998244353;
int inv[N], p[N], n, k, s;int main() {s = p[0] = inv[1] = 1;for (int i = 2; i <= 10000000; i++)inv[i] = MOD - 1ll * MOD / i * inv[MOD % i] % MOD;scanf("%d%d", &n, &k);for (int i = 1; i <= n; i++) {p[i] = 1ll * inv[i] * s % MOD;s = s + p[i];if (i >= k) s -= p[i - k];if (s >= MOD)s -= MOD;if (s < 0)s += MOD;}int ans = p[n];for (int i = 1; i <= n; i++)ans = 1ll * ans * i % MOD;printf("%d", ans);return 0;
}

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